#9 Geometry with Coordinates and Vectors

\(\overrightarrow{AB}=(a,b)\), \(\overrightarrow{AC}=(-b,a)\). Then \(AB^2=a^2+b^2=AC^2\), and \(\overrightarrow{AB}\cdot\overrightarrow{AC}=a(-b)+ba=0\). Hence \(AB=AC\) and \(AB\perp AC\).

\(\overrightarrow{PA}=(-1-x,-y)\), \(\overrightarrow{PB}=(1-x,-y)\). Their dot product is \((-1-x)(1-x)+y^2=x^2+y^2-1\). Perpendicularity is equivalent to this being zero, that is \(x^2+y^2=1\).

We have \(M\left(\frac m2,\frac n2\right)\). Then \(MA^2=\frac{m^2+n^2}{4}\), \(MB^2=\left(m-\frac m2\right)^2+\left(0-\frac n2\right)^2=\frac{m^2+n^2}{4}\), and similarly \(MC^2=\frac{m^2+n^2}{4}\). Thus all three distances are equal.

The midpoint of \(AC\) is \(\left(\frac{u+p}{2},\frac{v+q}{2}\right)\). The midpoint of \(BD\) is also \(\left(\frac{u+p}{2},\frac{v+q}{2}\right)\). They coincide, so the diagonals bisect each other.

Let the position vectors of \(A,B,C\) be \(a,b,c\). Then \(m=\frac{a+b}{2}\), \(n=\frac{a+c}{2}\). Hence \(\overrightarrow{MN}=n-m=\frac{c-b}{2}=\frac12\overrightarrow{BC}\). This gives both parallelism and the length ratio.

\(AC^2=(a+b)^2+h^2\). Also \(BD^2=(-a-b)^2+h^2=(a+b)^2+h^2\). Therefore \(AC=BD\).

\(CA^2=(u+1)^2+v^2\), \(CB^2=(u-1)^2+v^2\). Thus \(CA^2+CB^2=2u^2+2v^2+2=2CM^2+2\), since \(CM^2=u^2+v^2\).

Since \(AB\) is horizontal, the altitude from \(C\) has equation \(x=q\). Line \(AC\) has slope \(\frac rq\) when \(q\ne 0\), so the altitude from \(B\) has slope \(-\frac qr\) and equation \(y=-\frac qr(x-p)\). At \(x=q\), we get \(y=\frac{q(p-q)}{r}\). If \(q=0\), the same formula gives \(y=0\). Hence \(H\left(q,\frac{q(p-q)}{r}\right)\).

Substituting \(B(a,0)\) gives \(u=-a\), and substituting \(C(0,b)\) gives \(v=-b\). The circle is \(x^2+y^2-ax-by=0\). Completing squares gives \(\left(x-\frac a2\right)^2+\left(y-\frac b2\right)^2=\frac{a^2+b^2}{4}\). The centre is \(\left(\frac a2,\frac b2\right)\), the midpoint of \(BC\).

Let the position vectors of \(A,B,C,D\) be \(a,b,c,d\). Then \(p=\frac{a+b}{2}\), \(q=\frac{b+c}{2}\), \(r=\frac{c+d}{2}\), \(s=\frac{d+a}{2}\). We have \(\overrightarrow{PQ}=q-p=\frac{c-a}{2}\) and \(\overrightarrow{SR}=r-s=\frac{c-a}{2}\). Similarly, \(\overrightarrow{QR}=\overrightarrow{PS}\). Hence \(PQRS\) is a parallelogram.

We have \(P(t,0)\), \(Q(0,t)\). The vector \(\overrightarrow{PQ}=(-t,t)=t(-1,1)\), while \(\overrightarrow{BC}=(-1,1)\). Hence \(PQ\parallel BC\) and \(PQ=t\cdot BC\). Since \(t=\frac{m}{m+n}\), we get \(PQ:BC=m:(m+n)\).

The perpendicular bisector of \(AB\) has equation \(x=3\). From \(OA=OC\), we get \(9+y^2=(3-2)^2+(y-4)^2\), so \(y=1\). Thus \(O(3,1)\). By the orthocenter formula with \(p=6,q=2,r=4\), \(H(2,2)\). The centroid is \(G\left(\frac{0+6+2}{3},\frac{0+0+4}{3}\right)=\left(\frac83,\frac43\right)\). Vector \(\overrightarrow{OH}=(-1,1)\), and \(\overrightarrow{OG}=\left(-\frac13,\frac13\right)=\frac13\overrightarrow{OH}\). Therefore the points are collinear and \(OG:GH=1:2\).

Point \(M=(0,0)\). Since \(AB\) is horizontal, \(CM\perp AB\) if and only if \(u=0\). On the other hand, \(CA^2=(u+1)^2+v^2\), \(CB^2=(u-1)^2+v^2\). Equality of these squares is also equivalent to \(u=0\). Hence the conditions are equivalent.

Let \(P=B+t(C-B)=(6,0)+t(-4,5)\). Then \(\overrightarrow{AP}=P\), and \(\overrightarrow{BC}=(-4,5)\). The perpendicularity condition gives \(((6,0)+t(-4,5))\cdot(-4,5)=0\). Thus \(-24+41t=0\), so \(t=\frac{24}{41}\). Hence \(BP:PC=t:(1-t)=24:17\).

\(PA^2=(x+1)^2+y^2\), \(PB^2=(x-1)^2+y^2\). Their sum is \(2x^2+2y^2+2\). The condition \(PA^2+PB^2=10\) gives \(2x^2+2y^2+2=10\), that is \(x^2+y^2=4\). The locus is the circle with centre at the origin and radius \(2\).

By symmetry the centre should lie on the \(Oy\)-axis. Check the circle \(x^2+\left(y+\frac14\right)^2=\frac{145}{16}\). For \(A\) and \(B\): \(9+\frac{1}{16}=\frac{145}{16}\). For \(C\) and \(D\): \(4+\frac{81}{16}=\frac{145}{16}\). All four points satisfy one circle equation.

\(AB^2=BC^2=CD^2=DA^2=a^2+b^2\), so all sides are equal and \(ABCD\) is a rhombus. Diagonal \(AC\) lies on the \(Ox\)-axis, while \(BD\) lies on the \(Oy\)-axis, hence \(AC\perp BD\).

By segment division, \(D(4,2)\), \(E(0,\frac32)\), \(F(\frac{12}{5},0)\). Line \(AD\): \(y=\frac{x}{2}\). Line \(BE\): \(y=-\frac{x}{4}+\frac32\). Their intersection is \(x=2\), \(y=1\). Line \(CF\) passes through \(C(0,6)\) and \(F(\frac{12}{5},0)\), so it has equation \(y=6-\frac52x\). Point \((2,1)\) lies on it. Hence the three lines are concurrent.

The circle through \(A(0,0)\), \(B(1,0)\) has form \(x^2+y^2-x+\lambda y=0\). Substituting \(C(u,v)\), we get \(\lambda=\frac{u-u^2-v^2}{v}\). The reflected point is \(H'\left(u,-\frac{u(1-u)}{v}\right)\). Substitution gives \(u^2+\frac{u^2(1-u)^2}{v^2}-u-\frac{u(1-u)(u-u^2-v^2)}{v^2}=0\). After cancellation, this becomes \(0=0\). Hence \(H'\in(ABC)\).

Since \(D\) is the foot of the altitude from \(C\), \(CD\perp DB\), so \(\angle BDC=90^\circ\). Since \(E\) is the foot of the altitude from \(B\), \(BE\perp CE\), so \(\angle BEC=90^\circ\). Therefore points \(D\) and \(E\) lie on the circle with diameter \(BC\). Hence \(B,C,D,E\) are concyclic.

The circumcenter has coordinates \(O\left(\frac12,\frac{u^2+v^2-u}{2v}\right)\). The orthocenter is \(H\left(u,\frac{u(1-u)}{v}\right)\). The vector equality \(\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\) is equivalent to \(H=A+B+C-2O\). The right-hand side equals \((1+u,v)-\left(1,\frac{u^2+v^2-u}{v}\right)=\left(u,\frac{u-u^2}{v}\right)=H\). The equality is proved.

The centre is \(O(0,0)\), and the radius is \(5\). The foot of the perpendicular from \(O\) to the line \(3x+4y=20\) is \(M\left(\frac{3\cdot 20}{3^2+4^2},\frac{4\cdot 20}{3^2+4^2}\right)=\left(\frac{12}{5},\frac{16}{5}\right)\). The distance \(OM=4\). Half the chord is \(\sqrt{5^2-4^2}=3\), so \(AB=6\).

Choose affine coordinates so that \(E=(0,0)\), \(A=(a,0)\), \(B=(b,0)\), \(D=(0,d)\), \(C=(0,c)\). Let \(F=(r,s)\). Since \(F\in AD\), we have \(\frac{r}{a}+\frac{s}{d}=1\), so \(rd+as=ad\). Since \(F\in BC\), we have \(\frac{r}{b}+\frac{s}{c}=1\), so \(rc+bs=bc\). The midpoints of \(AC\), \(BD\), \(EF\) have coordinates \(\left(\frac a2,\frac c2\right)\), \(\left(\frac b2,\frac d2\right)\), \(\left(\frac r2,\frac s2\right)\). It remains to prove that \((r,s)\) lies on the line through \((a,c)\) and \((b,d)\). The equation of this line gives the expression \((d-c)r-(b-a)s-a(d-c)+c(b-a)\). It equals \((rd+as-ad)-(rc+bs-bc)=0\). Hence the three midpoints are collinear.

We have \(O\left(\frac12,\frac{u^2+v^2-u}{2v}\right)\), \(H\left(u,\frac{u(1-u)}{v}\right)\), \(G\left(\frac{1+u}{3},\frac v3\right)\). Check that \(G=\frac{2O+H}{3}\). For the first coordinate: \(\frac{2\cdot\frac12+u}{3}=\frac{1+u}{3}\). For the second coordinate: \(\frac{\frac{u^2+v^2-u}{v}+\frac{u-u^2}{v}}{3}=\frac{v}{3}\). Hence \(G\) lies on \(OH\) and divides segment \(OH\) in the ratio \(OG:GH=1:2\).